ABC26GN1371 · Square Roots and Cube Roots
Subject: General Aptitude · Chapter: Square Roots and Cube Roots · Exam: 2010 · Marks: · Difficulty:
If $\sqrt{x+\frac{x}{y}}=x \sqrt{\frac{x}{y}}$, where $x$ and $y$ are positive real numbers, then $y$ is equal to
(a)$x+1$
(b)$x-1$
(c)$x^{2}+1$
(d)$x^{2}-1$
Answer
Explanation
$$\begin{aligned} & \sqrt{x+\frac{x}{y}}=x \sqrt{\frac{x}{y}} \Rightarrow x+\frac{x}{y}=x^{2} \cdot \frac{x}{y} \Rightarrow \frac{x y+x}{y}=\frac{x^{3}}{y} \\ & \Rightarrow x y+x=x^{3} \\ & \Rightarrow y+1=x^{2} \\ & \Rightarrow y=x^{2}-1 \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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